Measure-Valued Variational Models with Applications to Diffusion-Weighted Imaging

Abstract

We develop a general mathematical framework for variational problems where the unknown function takes values in the space of probability measures on some metric space. We study weak and strong topologies and define a total variation seminorm for functions taking values in a Banach space. The seminorm penalizes jumps and is rotationally invariant under certain conditions. We prove existence of a minimizer for a class of variational problems based on this formulation of total variation and provide an example where uniqueness fails to hold. Employing the Kantorovich–Rubinstein transport norm from the theory of optimal transport, we propose a variational approach for the restoration of orientation distribution function-valued images, as commonly used in diffusion MRI. We demonstrate that the approach is numerically feasible on several data sets.

Appendices

Appendix A: Background from Functional Analysis and Measure Theory

In this appendix, we present the theoretical background for a rigorous understanding of the notation and definitions underlying the notion of ${\text {TV}}$ as proposed in (5) and (7). Section A.1 is concerned with Banach space-valued functions, and Sect. A.2 focuses on the special case of measure-valued functions.

1.1 Banach Space-Valued Functions of Bounded Variation

This subsection introduces a function space on which the formulation of ${\text {TV}}$ as given in (5) is well defined.

Let $(V, \Vert \cdot \Vert _V)$ be a real Banach space with (topological) dual space $V^*$, i.e., $V^*$ is the set of bounded linear operators from V to $\mathbb {R}$. The dual pairing is denoted by $\langle p, v \rangle := p(v)$ whenever $p \in V^*$ and $v \in V$.

We say that $u:\varOmega \rightarrow V$ is weakly measurable if $x \mapsto \langle p, u(x) \rangle$ is measurable for each $p \in V^*$ and say that $u \in L_w^\infty (\varOmega , V)$ if u is weakly measurable and essentially bounded in V, i.e.,

$$\begin{aligned} \Vert u\Vert _{\infty ,V} := \hbox {ess}\,\hbox {sup}_{x \in \varOmega } \Vert u(x)\Vert _V < \infty . \end{aligned}$$

(A.1)

Note that the essential supremum is well defined even for non-measurable functions as long as the measure is complete. In our case, we assume the Lebesgue measure on $\varOmega$ which is complete.

The following Lemma ensures that the integrand in (5) is measurable.

Lemma 1

Assume that $u:\varOmega \rightarrow V$ is weakly measurable and $p:\varOmega \rightarrow V^*$ is weakly* continuous, i.e., for each $v \in V$, the map $x \mapsto \langle p(x), v \rangle$ is continuous. Then, the map $x \mapsto \langle p(x), u(x) \rangle$ is measurable.

Proof

Define $f:\varOmega \times \varOmega \rightarrow \mathbb {R}$ via

$$\begin{aligned} f(x, \xi ) := \langle p(x), u(\xi ) \rangle . \end{aligned}$$

(A.2)

Then, f is continuous in the first and measurable in the second variable. In the calculus of variations, functions with this property are called Carathéodory functions and have the property that $x \mapsto f(x,g(x))$ is measurable whenever $g:\varOmega \rightarrow \varOmega$ is measurable, which is proven by approximation of g as the pointwise limit of simple functions [22, Proposition 3.7]. In our case, we can simply set $g(x) := x$, which is measurable, and the assertion follows. $\square$

1.2 Wasserstein Metrics and the KR Norm

This subsection is concerned with the definition of the space of measures $K\!R(X)$ and the isometric embedding $\mathcal {P}(X) \subset K\!R(X)$ underlying the formulation of ${\text {TV}}$ given in (7).

By $\mathcal {M}(X)$ and $\mathcal {P}(X) \subset \mathcal {M}(X)$, we denote the sets of signed Radon measures and Borel probability measures supported on X. $\mathcal {M}(X)$ is a vector space [40, p. 360] and a Banach space if equipped with the norm

$$\begin{aligned} \Vert \mu \Vert _\mathcal {M}:= \int _X \,\hbox {d}|\mu |, \end{aligned}$$

(A.3)

so that a function $u:\varOmega \rightarrow \mathcal {P}(X) \subset \mathcal {M}(X)$ is Banach space-valued (i.e., u takes values in a Banach space). If we define C(X) as the space of continuous functions on X with norm $\Vert f\Vert _{C} := \sup _{x \in X} |f(x)|$, under the above assumptions on X, $\mathcal {M}(X)$ can be identified with the (topological) dual space of C(X) with dual pairing

$$\begin{aligned} \langle \mu , p \rangle := \int _{X} p \,\hbox {d}\mu , \end{aligned}$$

(A.4)

whenever $\mu \in \mathcal {M}(X)$ and $p \in C(X)$, as proven in [40, p. 364]. Hence, $\mathcal {P}(X)$ is a bounded subset of a dual space.

We will now see that additionally, $\mathcal {P}(X)$ can be regarded as subset of a Banach space which is a predual space (in the sense that its dual space can be identified with a “meaningful” function space) and which metrizes the weak* topology of $\mathcal {M}(X)$ on $\mathcal {P}(X)$ by the optimal transport metrics we are interested in.

For $q \ge 1$, the Wasserstein metrics $W_q$ on $\mathcal {P}(X)$ are defined via

$$\begin{aligned} \begin{aligned} W_q(\mu , \mu ') :=&\left( \inf _{\gamma \in \varGamma (\mu ,\mu ')} \int _{X \times X} d(x,y)^q \,\hbox {d}\gamma (x,y) \right) ^{1/q}, \end{aligned} \end{aligned}$$

(A.5)

where

$$\begin{aligned} \varGamma (\mu ,\mu ') := \left\{ \gamma \in \mathcal {P}(X \times X):~\pi _1 \gamma = \mu , ~\pi _2 \gamma = \mu ' \right\} . \end{aligned}$$

(A.6)

Here, $\pi _i\gamma$ denotes the ith marginal of the measure $\gamma$ on the product space $X \times X$, i.e., $\pi _1\gamma (A) := \gamma (A \times X)$ and $\pi _2\gamma (B) := \gamma (X \times B)$ whenever $A, B \subset X$.

Now, let ${\text {Lip}}(X,\mathbb {R}^d)$ be the space of Lipschitz-continuous functions on X with values in $\mathbb {R}^d$ and ${\text {Lip}}(X) := {\text {Lip}}(X,\mathbb {R}^1)$. Furthermore, denote the Lipschitz seminorm by $[\cdot ]_{{\text {Lip}}}$ so that $[f]_{{\text {Lip}}}$ is the Lipschitz constant of f. Note that, if we fix some arbitrary $x_0 \in X$, the seminorm $[\cdot ]_{{\text {Lip}}}$ is actually a norm on the set

$$\begin{aligned} {\text {Lip}}_0(X,\mathbb {R}^d) := \{ p \in {\text {Lip}}(X,\mathbb {R}^d):p(x_0) = 0 \}. \end{aligned}$$

(A.7)

The famous Kantorovich–Rubinstein duality [44] states that, for $q=1$, the Wasserstein metric is actually induced by a norm, namely $W_1(\mu , \mu ') = \Vert \mu - \mu '\Vert _{K\!R}$, where

$$\begin{aligned} \Vert \nu \Vert _{K\!R} := \sup \left\{ \int _{X} p \,\hbox {d}\nu : ~p \in {\text {Lip}}_0(X), ~[p]_{{\text {Lip}}} \le 1 \right\} , \end{aligned}$$

(A.8)

whenever $\nu \in \mathcal {M}_0(X) := \{ \mu \in \mathcal {M}:\int _X d\mu = 0\}$. The completion $K\!R(X)$ of $\mathcal {M}_0(X)$ with respect to $\Vert \cdot \Vert _{K\!R}$ is a predual space of $({\text {Lip}}_0(X), [\cdot ]_{{\text {Lip}}})$ [79, Theorem 2.2.2 and Cor. 2.3.5].² Hence, after subtracting a point mass at $x_0$, the set $\mathcal {P}(X) - \delta _{x_0}$ is a subset of the Banach space $K\!R(X)$, the predual of ${\text {Lip}}_0(X)$.

Consequently, the embeddings

$$\begin{aligned} \mathcal {P}(X)&\hookrightarrow (K\!R(X), \Vert \cdot \Vert _{K\!R}), \end{aligned}$$

(A.9)

$$\begin{aligned} \mathcal {P}(X)&\hookrightarrow (\mathcal {M}(X), \Vert \cdot \Vert _{\mathcal {M}}) \end{aligned}$$

(A.10)

define two different topologies on $\mathcal {P}(X)$. The first embedding space $(\mathcal {M}(X), \Vert \cdot \Vert _{\mathcal {M}})$ is isometrically isomorphic to the dual of C(X). The second embedding space $(K\!R(X), \Vert \cdot \Vert _{K\!R})$ is known to be a metrization of the weak*-topology on the bounded subset $\mathcal {P}(X)$ of the dual space $\mathcal {M}(X) = C(X)^*$ [77, Theorem 6.9].

Importantly, while $(\mathcal {P}(X), \Vert \cdot \Vert _{\mathcal {M}})$ is not separable unless X is discrete, $(\mathcal {P}(X), \Vert \cdot \Vert _{K\!R})$ is in fact compact, in particular complete and separable [77, Theorem 6.18] which is crucial in our result on the existence of minimizers (Theorem 1).

Appendix B: Proof of ${\text {TV}}$-Behavior for Cartoonlike Functions

Proof

(Prop. 1) Let $p:\varOmega \rightarrow (V^*)^d$ satisfy the constraints in (5) and denote by $\nu$ the outer unit normal of $\partial U$. The set $\varOmega$ is bounded, p and its derivatives are continuous and $u \in L_w^\infty (\varOmega , V)$ since the range of u is finite and U, $\varOmega$ are measurable. Therefore, all of the following integrals converge absolutely. Due to linearity of the divergence,

$$\begin{aligned}&\langle {\text {div}}p(x), u^\pm \rangle = {\text {div}}(\langle p(\cdot ), u^\pm \rangle ), \end{aligned}$$

(B.1)

$$\begin{aligned}&\langle p(x), u^\pm \rangle := ( \langle p_1(x), u^\pm \rangle , \dots , \langle p_d(x), u^\pm \rangle ) \in \mathbb {R}^d. \end{aligned}$$

(B.2)

Using this property and applying Gauss’ theorem, we compute

$$\begin{aligned} \begin{aligned}&\int _\varOmega \langle -{\text {div}}p(x), u(x) \rangle \,\hbox {d}x \\&\quad = -\int _{\varOmega \setminus U} {\text {div}}(\langle p(x), u^- \rangle ) \,\hbox {d}x - \int _{U} {\text {div}}(\langle p(x), u^+ \rangle ) \,\hbox {d}x \\&\quad \overset{\text {Gauss}}{=} \int _{\partial U} \sum _{i=1}^d \langle \nu _i(x) p_i(x), {u^+} - {u^-} \rangle \,\hbox {d}\mathcal {H}^{d-1}(x) \\&\quad \le \mathcal {H}^{d-1}(\partial U) \cdot \Vert {u^+} - {u^-}\Vert _V. \end{aligned} \end{aligned}$$

(B.3)

For the last inequality, we used our first assumption on $\Vert \cdot \Vert _{(V^*)^d}$ together with the norm constraint for p in (5). Taking the supremum over p as in (5), we arrive at

$$\begin{aligned} {\text {TV}}_{V}(u) \le \mathcal {H}^{d-1}(\partial U) \cdot \Vert {u^+} - {u^-}\Vert _V. \end{aligned}$$

(B.4)

For the reverse inequality, let ${\tilde{p}} \in V^*$ be arbitrary with the property $\Vert {\tilde{p}}\Vert _{V^*} \le 1$ and $\phi \in C_c^1(\varOmega , \mathbb {R}^d)$ satisfying $\Vert \phi (x)\Vert _2 \le 1$. Now, by (11), the function

$$\begin{aligned} p(x) := (\phi _1(x) {\tilde{p}}, \dots , \phi _d(x) {\tilde{p}}) \in (V^*)^d \end{aligned}$$

(B.5)

has the properties required in (5). Hence,

$$\begin{aligned} {\text {TV}}_{V}(u)&\ge \int _\varOmega \langle -{\text {div}}p(x), u(x) \rangle \,\hbox {d}x \end{aligned}$$

(B.6)

$$\begin{aligned}&= -\int _\varOmega {\text {div}}\phi (x) \,\hbox {d}x \cdot \langle {\tilde{p}}, u^+ - u^- \rangle . \end{aligned}$$

(B.7)

Taking the supremum over all $\phi \in C_c^1(\varOmega , \mathbb {R}^d)$ satisfying $\Vert \phi (x)\Vert _2 \le 1$, we obtain

$$\begin{aligned} {\text {TV}}_{V}(u)&\ge {\text {Per}}(U, \varOmega ) \cdot \langle {\tilde{p}}, u^+ - u^- \rangle , \end{aligned}$$

(B.8)

where ${\text {Per}}(U, \varOmega )$ is the perimeter of U in $\varOmega$. In the theory of Caccioppoli sets (or sets of finite perimeter), the perimeter is known to agree with $\mathcal {H}^{d-1}(\partial U)$ for sets with $C^1$ boundary [4, p. 143].

Now, taking the supremum over all ${\tilde{p}} \in V^*$ with $\Vert {\tilde{p}}\Vert _{V^*} \le 1$ and using the fact that the canonical embedding of a Banach space into its bidual is isometric, i.e.,

$$\begin{aligned} \Vert u\Vert _V = \sup _{\Vert p\Vert _{V^*} \le 1} \langle p, u \rangle , \end{aligned}$$

(B.9)

we arrive at the desired reverse inequality which concludes the proof. $\square$

Appendix C: Proof of Rotational Invariance

Proof

(Proposition 2) Let $R \in SO(d)$ and define

$$\begin{aligned} R^T\varOmega := \{ R^T x : x \in \varOmega \}, ~{\tilde{p}}(y) := R^T p(Ry). \end{aligned}$$

(C.1)

In (5), the norm constraint on p(x) is equivalent to the norm constraint on ${\tilde{p}}(y)$ by condition (13). Now, consider the integral transform

$$\begin{aligned} \int _\varOmega \langle -{\text {div}}p(x), u(x) \rangle \,\hbox {d}x&= \int _{R^T\varOmega } \langle -{\text {div}}p(R y), {\tilde{u}}(y) \rangle \,\hbox {d}y \end{aligned}$$

(C.2)

$$\begin{aligned}&= \int _{R^T\varOmega } \langle -{\text {div}}{\tilde{p}}(y), {\tilde{u}}(y) \rangle \,\hbox {d}y. \end{aligned}$$

(C.3)

where, using $R^T R = I$,

$$\begin{aligned} {\text {div}}{\tilde{p}}(y)&= \sum _{i=1}^d \partial _i {\tilde{p}}_i(y) = \sum _{i=1}^d \sum _{j=1}^d R_{ji} \partial _i \left[ p_j(R y)\right] \end{aligned}$$

(C.4)

$$\begin{aligned}&= \sum _{i=1}^d \sum _{j=1}^d \sum _{k=1}^d R_{ji} R_{ki} \partial _k p_j(R y) \end{aligned}$$

(C.5)

$$\begin{aligned}&= \sum _{j=1}^d \sum _{k=1}^d \partial _k p_j(R y) \sum _{i=1}^d R_{ji} R_{ki} \end{aligned}$$

(C.6)

$$\begin{aligned}&= \sum _{j=1}^d \partial _j p_j(R y) = {\text {div}}p(R y), \end{aligned}$$

(C.7)

which implies ${\text {TV}}_V(u) = {\text {TV}}_V({\tilde{u}})$. $\square$

Appendix D: Discussion of Product Norms

There is one subtlety about formulation (5) of the total variation: The choice of norm for the product space $(V^*)^d$ affects the properties of our total variation seminorm.

1.1 Product Norms as Required in Proposition 1

The following proposition gives some examples for norms that satisfy or fail to satisfy conditions (10) and (11) in Proposition 1 about cartoonlike functions.

Proposition 4

The following norms for $p \in (V^*)^d$ satisfy (10) and (11) for any normed space V:

• 1. For $s = 2$:

  $$\begin{aligned} \Vert p\Vert _{(V^*)^d,s} := \left( \sum _{i=1}^d \Vert p_i\Vert _{V^*}^s \right) ^{1/s}. \end{aligned}$$

  (D.1)

• 2. Writing $p(v):=(\langle p_1,v\rangle ,\dots ,\langle p_d,v\rangle )\in \mathbb {R}^d$, $v \in V$,

  $$\begin{aligned} \Vert p\Vert _{\mathcal {L}(V, \mathbb {R}^d)} := \sup _{\Vert v\Vert _V \le 1} \Vert p(v)\Vert _{2} \end{aligned}$$

  (D.2)

On the other hand, for any $1 \le s < 2$ and $s > 2$, there is a normed space V such that at least one of the properties (10), (11) is not satisfied by corresponding product norm (D.1).

Remark 1

In the finite-dimensional Euclidean case $V = \mathbb {R}^n$ with norm $\Vert \cdot \Vert _2$, we have $(V^*)^d = \mathbb {R}^{d,n}$; thus, p is matrix-valued and $\Vert \cdot \Vert _{\mathcal {L}(V, \mathbb {R}^d)}$ agrees with the spectral norm $\Vert \cdot \Vert _\sigma$. The norm defined in (D.1) is the Frobenius norm $\Vert \cdot \Vert _F$ for $s=2$.

Proof

(Prop. 4) By Cauchy–Schwarz,

$$\begin{aligned} \left| \textstyle {\sum _{i=1}^d} x_i \langle p_i, v \rangle \right|&\le \Vert x\Vert _2 \left( \textstyle {\sum _{i=1}^d} \left| \langle p_i, v \rangle \right| ^2 \right) ^{1/2} \end{aligned}$$

(D.3)

$$\begin{aligned}&\le \Vert x\Vert _2 \left( \textstyle {\sum _{i=1}^d} \Vert p_i\Vert _{V^*}^2 \Vert v\Vert _V^2 \right) ^{1/2} \end{aligned}$$

(D.4)

$$\begin{aligned}&\le \Vert x\Vert _2 \Vert v\Vert _V \left( \textstyle {\sum _{i=1}^d} \Vert p_i\Vert _{V^*}^2 \right) ^{1/2}, \end{aligned}$$

(D.5)

whenever $p \in (V^*)^d$, $v \in V$, and $x \in \mathbb {R}^d$. Similarly, for each $q \in V^*$,

$$\begin{aligned} \left( \textstyle {\sum _{i=1}^d} \Vert x_i q\Vert _{V^*}^2 \right) ^{1/2} = \Vert x\Vert _2 \Vert q\Vert _{V^*}. \end{aligned}$$

(D.6)

Hence, for $s = 2$, properties (10) and (11) are satisfied by product norm (D.1).

For operator norm (D.2), consider

$$\begin{aligned} \left| \textstyle {\sum _{i=1}^d} x_i \langle p_i, v \rangle \right|&\le \Vert x\Vert _2 \left( \textstyle {\sum _{i=1}^d} \left| \langle p_i, v \rangle \right| ^2 \right) ^{1/2} \end{aligned}$$

(D.7)

$$\begin{aligned}&= \Vert x\Vert _2 \Vert p(v)\Vert _2 \end{aligned}$$

(D.8)

$$\begin{aligned}&\le \Vert x\Vert _2 \Vert p\Vert _{\mathcal {L}(V, \mathbb {R}^d)} \Vert v\Vert _V, \end{aligned}$$

(D.9)

which is property (10). On the other hand, (11) follows from

$$\begin{aligned} \Vert (x_1 q, \dots , x_d q)\Vert _{\mathcal {L}(V, \mathbb {R}^d)}&= \sup _{\Vert v\Vert _V \le 1} \left( \textstyle {\sum _{i=1}^d} |x_i q(v)|^2 \right) ^{1/2} \end{aligned}$$

(D.10)

$$\begin{aligned}&= \Vert x\Vert _2 \sup _{\Vert v\Vert _V \le 1} |q(v)| \end{aligned}$$

(D.11)

$$\begin{aligned}&= \Vert x\Vert _2 \Vert q\Vert _{V^*}. \end{aligned}$$

(D.12)

Now, for $s > 2$, property (10) fails for $d = 2$, $V = V^* = \mathbb {R}$, $p = x = (1,1)$ and $v = 1$ since

$$\begin{aligned} \left| \sum _{i=1}^d x_i \langle p_i, v \rangle \right|&= 2 > 2^{1/2} \cdot 2^{1/s} = \Vert x\Vert _2 \Vert p\Vert _{(V^*)^d,s} \Vert v\Vert _V. \end{aligned}$$

(D.13)

For $1 \le s < 2$, consider $d = 2$, $V^* = \mathbb {R}$, $q = 1$ and $x = (1,1)$, then

$$\begin{aligned} \Vert (x_1 q, \dots , x_d q)\Vert _{(V^*)^d,s}&= 2^{1/s} > 2^{1/2} = \Vert x\Vert _2 \Vert q\Vert _{V^*}, \end{aligned}$$

(D.14)

which contradicts property (11). $\square$

1.2 Rotationally Symmetric Product Norms

For $V = (\mathbb {R}^n, \Vert \cdot \Vert _2)$, property (13) in Proposition 2 is satisfied by the Frobenius norm as well as the spectral norms on $(V^*)^d = \mathbb {R}^{d,n}$. In general, the following proposition holds:

Proposition 5

For any normed space V, rotational invariance property (13) is satisfied by operator norm (D.2). For any $s \in [1,\infty )$, there is a normed space V such that property (13) does not hold for product norm (D.1).

Proof

By definition of the operator norm and rotational invariance of the Euclidean norm $\Vert \cdot \Vert _2$,

$$\begin{aligned} \Vert Rp\Vert _{\mathcal {L}(V, \mathbb {R}^d)}&= \sup _{\Vert v\Vert _V \le 1} \Vert Rp(v)\Vert _{2} \end{aligned}$$

(D.15)

$$\begin{aligned}&= \sup _{\Vert v\Vert _V \le 1} \Vert p(v)\Vert _{2} = \Vert p\Vert _{\mathcal {L}(V, \mathbb {R}^d)}. \end{aligned}$$

(D.16)

For product norms (D.1), without loss of generality, we consider the case $d = 2$, $V := (\mathbb {R}^2, \Vert \cdot \Vert _1)$, $p_1 = (1,0)$, $p_2 = (0,1)$ and

$$\begin{aligned} R := \begin{pmatrix} 1/2 &{} -\sqrt{3}/2 \\ \sqrt{3}/2 &{} 1/2 \end{pmatrix} \in SO(2). \end{aligned}$$

(D.17)

Then, $V^* := (\mathbb {R}^2, \Vert \cdot \Vert _\infty )$ and

$$\begin{aligned} \Vert p\Vert _{(V^*)^d,s} = \left( \textstyle {\sum _{i=1}^2} \Vert p_i\Vert _\infty ^s \right) ^{1/s} = 2^{1/s} \end{aligned}$$

(D.18)

whereas

$$\begin{aligned} (Rp)_1&= (1/2, -\sqrt{3}/2), ~(Rp)_2 = (\sqrt{3}/2, 1/2), \end{aligned}$$

(D.19)

$$\begin{aligned} \Vert Rp\Vert _{(V^*)^d,s}&= \left( \textstyle {\sum _{i=1}^2} (\sqrt{3}/2)^s \right) ^{1/s} \end{aligned}$$

(D.20)

$$\begin{aligned}&= 2^{1/s} \cdot \sqrt{3}/2 \ne 2^{1/s} = \Vert p\Vert _{(V^*)^d,s}, \end{aligned}$$

(D.21)

for any $1 \le s < \infty$. $\square$

1.3 Product Norms on ${\text {Lip}}_0(X)$

We conclude our discussion about product norms on $(V^*)^d$ with the special case of $V = K\!R(X)$: For $p \in [{\text {Lip}}_0(X)]^d$, the most natural choice is

$$\begin{aligned}{}[p]_{{\text {Lip}}(X,\mathbb {R}^d)} := \sup _{z \ne z'} \frac{\Vert p(z) - p(z')\Vert ^2_2}{d(z,z')}, \end{aligned}$$

(D.22)

which is automatically rotationally invariant. On the other hand, the product norm defined in (D.1) (with $s=2$), namely $\sqrt{\sum _{i=1}^d [p_i]_{{\text {Lip}}}^2}$, is not rotationally invariant for general metric spaces X. However, in the special case $X \subset (\mathbb {R}^n, \Vert \cdot \Vert _2)$ and $p \in C^1(X,\mathbb {R}^d)$, norms (D.22) and (D.1) coincide with $\sup _{z\in X} \Vert Dp(z)\Vert _\sigma$ (spectral norm of the Jacobian) and $\sup _{z\in X} \Vert Dp(z)\Vert _F$ (Frobenius norm of the Jacobian), respectively, both satisfying rotational invariance.

Appendix E: Proof of Non-uniqueness

Proof

(Prop. 3) Let $u \in L_w^\infty (\varOmega , \mathcal {P}(X))$. With the given choice of X, there exists a measurable function ${\tilde{u}}:\varOmega \rightarrow [0,1]$ such that

$$\begin{aligned} u(x) = {\tilde{u}}(x) \delta _0 + (1 - {\tilde{u}}(x)) \delta _1. \end{aligned}$$

(E.1)

The measurability of ${\tilde{u}}$ is equivalent to the weak measurability of u by definition:

$$\begin{aligned} \langle p, u(x) \rangle&= {\tilde{u}}(x) \cdot p_0 + (1 - {\tilde{u}}(x)) \cdot p_1 \end{aligned}$$

(E.2)

$$\begin{aligned}&= {\tilde{u}}(x) \cdot (p_0 - p_1) + p_1. \end{aligned}$$

(E.3)

The constraint

$$\begin{aligned} p \in C_c^1(\varOmega , [{\text {Lip}}_0(X)]^d), ~[p(x)]_{{\text {Lip}}(X,\mathbb {R}^d)} \le 1 \end{aligned}$$

(E.4)

from the definition of ${\text {TV}}_{K\!R}$ in (7) translates to

$$\begin{aligned} p_0, p_1 \in C_c(\varOmega ,\mathbb {R}^d),~\Vert p_0(x) - p_1(x)\Vert _2 \le 1. \end{aligned}$$

(E.5)

Furthermore,

$$\begin{aligned}&\langle -{\text {div}}p(x), u(x) \rangle \end{aligned}$$

(E.6)

$$\begin{aligned}&\quad = -{\text {div}}p_0(x) \cdot {\tilde{u}}(x) - {\text {div}}p_1(x) \cdot (1-{\tilde{u}}(x)) \end{aligned}$$

(E.7)

$$\begin{aligned}&\quad = -{\text {div}}(p_0 - p_1)(x) \cdot {\tilde{u}}(x) - {\text {div}}p_1(x). \end{aligned}$$

(E.8)

By the compact support of $p_1$, the last term vanishes when integrated over $\varOmega$. Consequently,

$$\begin{aligned}&{\text {TV}}_{K\!R}(u) = \sup \left\{ \int _\varOmega -{\text {div}}(p_0 - p_1) (x) \cdot {\tilde{u}}(x) \,\hbox {d}x : \right. \end{aligned}$$

(E.9)

$$\begin{aligned}&\qquad \left. p_0, p_1 \in C_c(\varOmega ,\mathbb {R}^d),~\Vert (p_0 - p_1)(x)\Vert _2 \le 1 \right\} \end{aligned}$$

(E.10)

$$\begin{aligned}&\quad = \sup \left\{ \int _\varOmega -{\text {div}}p(x) \cdot {\tilde{u}}(x) \,\hbox {d}x : \right. \end{aligned}$$

(E.11)

$$\begin{aligned}&\qquad \left. p \in C_c(\varOmega ,\mathbb {R}^{d}),~\Vert p(x)\Vert _2 \le 1 \right\} \end{aligned}$$

(E.12)

$$\begin{aligned}&\quad = {\text {TV}}({\tilde{u}}). \end{aligned}$$

(E.13)

and therefore

$$\begin{aligned} T_{\rho ,\lambda }(u)&= \int _{\varOmega \setminus U} {\tilde{u}}(x) \,\hbox {d}x + \int _U (1-{\tilde{u}}(x)) \,\hbox {d}x + \lambda {\text {TV}}({\tilde{u}}) \end{aligned}$$

(E.14)

$$\begin{aligned}&= \int _\varOmega |\mathbf {1}_U(x) - {\tilde{u}}(x)| \,\hbox {d}x + \lambda {\text {TV}}({\tilde{u}}) \end{aligned}$$

(E.15)

$$\begin{aligned}&= \Vert \mathbf {1}_U - {\tilde{u}}\Vert _{L^1} + \lambda {\text {TV}}({\tilde{u}}). \end{aligned}$$

(E.16)

Thus we have shown that the functional $T_{\rho ,\lambda }$ is equivalent to the classical $L^1$-${\text {TV}}$ functional with the indicator function $\mathbf {1}_U$ as input data and evaluated at ${\tilde{u}}$ which is known to have non-unique minimizers for a certain choice of $\lambda$ [17]. $\square$

Appendix F: Proof of Existence

1.1 Well-Defined Energy Functional

In order for the functional defined in (15) to be well defined, the mapping $x \mapsto \rho (x, u(x))$ needs to be measurable. In the following lemma, we show that this is the case under mild conditions on $\rho$.

Lemma 2

Let $\rho :\varOmega \times \mathcal {P}(X) \rightarrow [0,\infty )$ be a globally bounded function that is measurable in the first and convex in the second variable, i.e., $x \mapsto \rho (x,\mu )$ is measurable for each $\mu \in \mathcal {P}(X)$, and $\mu \mapsto \rho (x,\mu )$ is convex for each $x \in \varOmega$. Then, the map $x \rightarrow \rho (x,u(x))$ is measurable for every $u \in L_w^\infty (\varOmega , \mathcal {P}(X))$.

Remark 2

As will become clear from the proof, the convexity condition can be replaced by the assumption that $\rho$ be continuous with respect to $(\mathcal {P}(X), W_1)$ in the second variable. However, in order to ensure weak* lower semicontinuity of functional (15), we will require convexity of $\rho$ in the existence proof (Theorem 1) anyway. Therefore, for simplicity we also stick to the (stronger) convexity condition in Lemma 2.

Remark 3

One example of a function satisfying the assumptions in Lemma 2 is given by

$$\begin{aligned} \rho (x,\mu ) := W_1(f(x),\mu ), ~x \in \varOmega , ~\mu \in \mathcal {P}(\mathbb {S}^2). \end{aligned}$$

(F.1)

Indeed, boundedness follows from the boundedness of the Wasserstein metric in the case of an underlying bounded metric spaces (here $\mathbb {S}^2$). Convexity in the second argument follows from the fact that the Wasserstein metric is induced by a norm (A.8).

Proof

(Lemma 2) The metric space $(\mathcal {P}(X), W_1)$ is compact, hence separable. By Pettis’ measurability theorem [10, Chapter VI, §1, No. 5, Proposition 12], weak and strong measurability coincide for separably valued functions, so that u is actually strongly measurable as a function with values in $(\mathcal {P}(X),W_1)$. Note, however, that this does not imply strong measurability with respect to the norm topology of $(\mathcal {M}(X), \Vert \cdot \Vert _{\mathcal {M}})$ in general!

As bounded convex functions are locally Lipschitz continuous [19, Theorem 2.34], $\rho$ is continuous in the second variable with respect to $W_1$. As in the proof of Lemma 1, we now note that $\rho$ is a Carathéodory function, for which compositions with measurable functions such as $x \mapsto \rho (x,u(x))$ are known to be measurable. $\square$

1.2 The Notion of Weakly* Measurable Functions

Before we can go on with the proof of existence of minimizers to (15), we introduce the notion of weak* measurability because this will play a crucial role in the proof.

Analogously with the notion of weak measurability and with $L_{w}^\infty (\varOmega , K\!R(X))$ introduced above, we say that a measure-valued function $u:\varOmega \rightarrow \mathcal {M}(X)$ is weakly* measurable if the mapping

$$\begin{aligned} x \mapsto \int _X f(z) \,\hbox {d}u_x(z) \end{aligned}$$

(F.2)

is measurable for each $f \in C(X)$. $L_{w*}^\infty (\varOmega , \mathcal {M}(X))$ is defined accordingly as the space of weakly* measurable functions.

For functions $u:\varOmega \rightarrow \mathcal {P}(X)$ mapping onto the space of probability measures, there is an immediate connection between weak* measurability and weak measurability: u is weakly measurable if the mapping

$$\begin{aligned} x \mapsto \int _X p(z) \,\hbox {d}u_x(z) \end{aligned}$$

(F.3)

is measurable whenever $p \in {\text {Lip}}_0(X)$. However, since, by the Stone–Weierstrass theorem, the Lipschitz functions ${\text {Lip}}(X)$ are dense in $(C(X), \Vert \cdot \Vert _{\infty })$ [13, p. 198], both notions of measurability coincide for probability measure-valued functions $u:\varOmega \rightarrow \mathcal {P}(X)$, so that

$$\begin{aligned} L_w^\infty (\varOmega , \mathcal {P}(X)) = L_{w*}^\infty (\varOmega , \mathcal {P}(X)). \end{aligned}$$

(F.4)

However, as this equivalence does not hold for the larger spaces $L_{w*}^\infty (\varOmega , \mathcal {M}(X))$ and $L_{w}^\infty (\varOmega , \mathcal {M}(X))$, it will be crucial to keep track of the difference between weak and weak* measurability in the existence proof.

1.3 Proof of Existence

Proof

(Theorem 1) The proof is guided by the direct method from the calculus of variations. The first part is inspired by the proof of the fundamental theorem for Young measures as formulated and proven in [6].

Let $u^k:\varOmega \rightarrow \mathcal {P}(X)$, $k \in \mathbb {N}$, be a minimizing sequence for $T_{\rho ,\lambda }$, i.e.,

$$\begin{aligned} T_{\rho ,\lambda }(u^k) \rightarrow \inf _{u} T_{\rho ,\lambda }(u) ~\text { as }~ k \rightarrow \infty . \end{aligned}$$

(F.5)

As $\mathcal {M}(X)$ is the dual space of C(X), $L_{w*}^\infty (\varOmega , \mathcal {M}(X))$ with the norm defined in (A.1) is dual to the Banach space $L^1(\varOmega , C(X))$ of Bochner integrable functions on $\varOmega$ with values in C(X) [42, p. 93]. Now, $\mathcal {P}(X)$ as a subset of $\mathcal {M}(X)$ is bounded so that our sequence $u^k$ is bounded in $L_{w*}^\infty (\varOmega , \mathcal {M}(X))$ (here we use again that $L_{w*}^\infty (\varOmega , \mathcal {P}(X)) = L_{w}^\infty (\varOmega , \mathcal {P}(X))$).

Note that we get boundedness of our minimizing sequence “for free”, without any assumptions on the coercivity of $T_{\rho ,\lambda }$! Hence we can apply the Banach–Alaoglu theorem, which states that there exist $u^\infty \in L_{w*}^\infty (\varOmega , \mathcal {M}(X))$ and a subsequence, also denoted by $u^k$, such that

$$\begin{aligned} u^k \overset{*}{\rightharpoonup } u^\infty \text { in } L_{w*}^\infty (\varOmega , \mathcal {M}(X)). \end{aligned}$$

(F.6)

Using the notation in (A.4), this means by definition

$$\begin{aligned}&\int _\varOmega \langle u^k(x), p(x) \rangle \,\hbox {d}x \rightarrow \int _\varOmega \langle u^\infty (x), \,p(x) \rangle \,\hbox {d}x \end{aligned}$$

(F.7)

$$\begin{aligned}&\quad \forall p \in L^1(\varOmega , C(X)). \end{aligned}$$

(F.8)

We now show that $u^\infty (x) \in \mathcal {P}(X)$ almost everywhere, i.e., $u^\infty$ is a nonnegative measure of unit mass: Convergence (F.7) holds in particular for the choice $p(x,s) := \phi (x)f(s)$, where $\phi \in L^1(\varOmega )$ and $f \in C(X)$. For nonnegative functions $\phi$ and f, we have

$$\begin{aligned} \int _\varOmega \phi (x) \langle u^k(x), f \rangle \,\hbox {d}x \ge 0 \end{aligned}$$

(F.9)

for all k, which implies

$$\begin{aligned} \int _\varOmega \phi (x) \langle u^\infty (x), f \rangle \,\hbox {d}x \ge 0. \end{aligned}$$

(F.10)

Since this holds for all nonnegative $\phi$ and f, we deduce that $u^\infty (x)$ is a nonnegative measure for almost every $x \in \varOmega$. The choice $f(s) \equiv 1$ in (F.7) shows that $u^\infty$ has unit mass almost everywhere.

Therefore, $u^\infty (x) \in \mathcal {P}(X)$ almost everywhere and we have shown that $u^\infty$ lies in the feasible set $L_{w}^\infty (\varOmega , \mathcal {P}(X))$. It remains to show that $u^\infty$ is in fact a minimizer.

In order to do so, we prove weak* lower semicontinuity of $T_{\rho ,\lambda }$. We consider the two integral terms in definition (15) of $T_{\rho ,\lambda }$ separately. For the ${\text {TV}}_{K\!R}$ term, for any $p \in C_c^1(\varOmega , {\text {Lip}}(X,\mathbb {R}^d))$, we have ${\text {div}}p \in L^1(\varOmega , C(X))$ so that

$$\begin{aligned} \lim _{k\rightarrow \infty } \int _\varOmega \langle {\text {div}}u^k(x), p(x) \rangle \,\hbox {d}x = \int _\varOmega \langle {\text {div}}u^\infty (x), p(x) \rangle \,\hbox {d}x. \end{aligned}$$

(F.11)

Taking the supremum over all p with $[p(x)]_{[{\text {Lip}}(X)]^d} \le 1$ almost everywhere, we deduce lower semicontinuity of the regularizer:

$$\begin{aligned} {\text {TV}}_{K\!R}(u^\infty ) \le \liminf _{k\rightarrow \infty } {\text {TV}}_{K\!R}(u^k). \end{aligned}$$

(F.12)

The data fidelity term $u \mapsto \int _\varOmega \rho (x,u(x)) \,\hbox {d}x$ is convex and bounded on the closed convex subset $L_w^\infty (\varOmega , \mathcal {P}(X))$ of the space $L_{w*}^\infty (\varOmega , \mathcal {M}(X))$. It is also continuous, as convex and bounded functions on normed spaces are locally Lipschitz continuous. This implies weak* lower semicontinuity on $L_w^\infty (\varOmega , \mathcal {P}(X))$.

Therefore, the objective function $T_{\rho ,\lambda }$ is weakly* lower semicontinuous, and we obtain

$$\begin{aligned} T_{\rho ,\lambda }(u^\infty ) \le \lim \inf _{k\rightarrow \infty } T_{\rho ,\lambda }(u^k) \end{aligned}$$

(F.13)

for the minimizing sequence $(u^k)$, which concludes the proof.

$\square$